atticus30
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The height of a small rock falling from the top of a 124-ft-tall building with an initial downward velocity of [tex]$-30 \, \text{ft/sec}$[/tex] is modeled by the equation [tex]$h(t)=-16 t^2-30 t+124$[/tex], where [tex][tex]$t$[/tex][/tex] is the time in seconds. For which interval of time does the rock remain in the air?

A. [tex]t=2[/tex]
B. [tex]-2\ \textless \ t\ \textless \ 0[/tex]
C. [tex]0\ \textless \ t\ \textless \ 2[/tex]
D. [tex]t\ \textgreater \ 2[/tex]

Sagot :

GinnyAnswer
To determine the interval of time during which the rock remains in the air, we need to find the time [tex]\( t \)[/tex] at which the rock hits the ground. The height [tex]\( h(t) \)[/tex] at ground level is 0. Therefore, we need to solve the equation:

[tex]\[ h(t) = -16t^2 - 30t + 124 = 0 \][/tex]

The solutions to this equation will give us the times at which the rock is at ground level. By examining the problem, we are looking for the positive solution since time must be positive.

By solving the quadratic equation, we find:

[tex]\[ t = 2 \][/tex]

This indicates that the rock hits the ground at [tex]\( t = 2 \)[/tex] seconds. Since the rock remains in the air from the time it is released until the time it hits the ground, we can conclude that the interval of time during which the rock remains in the air is:

[tex]\[ 0 < t < 2 \][/tex]

Thus, the correct interval of time is:

[tex]\[ 0 < t < 2 \][/tex]
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